Answer
$r^{'}(f^4)= 12f^3[3\cos f^4+ 1]$
Work Step by Step
$r(m)=9\sin m+3m; m(f)=f^4$
Composite function will be:
$r(f^4)=9\sin f^4+3f^4$
Taking derivative with respect to f
$r^{'}(f^4)= \frac{d( 9\sin f^4+3f^4 )}{df}$
$r^{'}(f^4)= \frac{d( 9\sin f^4 )}{df}+ \frac{d(3f^4)}{df}$
$r^{'}(f^4)= 9\frac{d( \sin f^4 )}{df}+ 3 \frac{d(f^4)}{df}$
$r^{'}(f^4)= 9(\cos f^4)\frac{df^4}{df}+ 3 (4f^3)$
$r^{'}(f^4)= 9(\cos f^4)(4f^3)+ 3 (4f^3)$
$r^{'}(f^4)= 36f^3(\cos f^4)+ 12f^3$
$r^{'}(f^4)= 12f^3[3\cos f^4+ 1]$
